129 On Wilkinson's Method of treating the Nine-Points Circle, with Generalizations. By R. F. MUIRHBAD, M.A., D.Sc. (Bead 12th December 1909. Received 25th April 1910). Ever since I first became acquainted with Wilkinson's method of establishing the existence of the nine-points circle of a triangle (see Mackay's " Euclid," Appendix to Bk. IV., Prop. 2, the lettering of which I have followed in the first three sections of this paper), its simple and fundamental character has pleased me. I propose to point out first that this method yields probably the most elementary proof of the concurrence of the perpendiculars from the vertices, and then, after restating the investigation of the nine-points circle, to sketch some generalizations. §1- The method depends on the following propositions, which are either contained in Euclid's first book or are easy and well- known deductions from it. I shall use the symbol = to denote " is equal and parallel to, and has the same sense as." Prop, (i) If AB = CD then AC = BD. „ (ii) If AB || CD and AC || BD then AB = CD and AC = BD. „ (iii) If AB = CD then AD and BC bisect one another. „ (iv) If AD and CD bisect one another, then AB = CD and ACsBD. „ (v) If D is the mid point of AB and E that of AC, then BC = 2DE. „ (vi) If D is the mid point of AB and DE || BC, E being in AC, then E is the mid point of AC. , (vii) If ABCD is a rectangle, AC and BD are equal and bisect one another. „ (viii) If AC and BD are equal and bisect one another, ABCD is a rectangle. 10 Vol.28 Downloaded from https://www.cambridge.org/core. 01 Oct 2021 at 06:15:37, subject to the Cambridge Core terms of use . 130 Let ABC be a triangle (Fig. \). Let the perpendiculars BY, CZ let fall from B and C on AC and AB intersect in 0. Pig. 1. Let AO meet BC in X: we navel to prove that AX is per- pendicular to BC. Let H, K, L, be the mid points of BC, CA, AB respectively. „ TJ.V.W, „ „ „ OA,OB,OC By (v) Hence TJVHK is a parallelogram whose sides are parallel to AB and CO, which are at right angles. Hence "UVHK. is a rectangle. Hence by (vii) HTT and KV are equal, and bisect one another. Similarly WULH is a rectangle, and HU and LW are equal, and bisect one another. Hence BLV and LW are equal, and bisect one another. Hence KLVW is a rectangle by (viii). Hence BC and AX, which by (v) are parallel to its sides, are mutually perpendicular. Thus the three perpendiculars from A, B, C on the opposite sides are concurrent. Q. E. D. Downloaded from https://www.cambridge.org/core. 01 Oct 2021 at 06:15:37, subject to the Cambridge Core terms of use . 131 Cor. 1. By (vii) HXJ, KV, LW are all equal and bisect one another in the same point M, which is therefore the centre of a circle passing through XJVWHKL. By (vii) this circle also passes through XYZ, since HXU form two sides of a rectangle whose mid point is M. Cor. 2. By (v) OVHW, OWKU, OIJLV are parallelograms, and if we complete the parallelogram LVHS, then SH = LV = UO = KW by (ii). Hence by (iii) M is the mid point of SO. By (i) SK Hence 8 is the circumcentre. Also AO = 2UO = SH, etc. Cor. S. The triangles UVW and ABC are homothetic, O being the homothetic centre, and M and S corresponding points, the ratio of linear dimensions being 1 : 2. Hence OX produced meets the circumcircle of ABC at a distance from O = 2OX. Cor. 4. If AH meets SO in G, since A0 = 2SH, OG = 2GS. Similarly BK and CL cut OS in the same point G, which is there- fore the point of concurrence of the medians. §2- The generalization from the nine-points circle to the nine- points conic worked out by Dr P. Pinkerton in his paper in Vol. XXIV. can be established with great ease by Wilkinson's method. Let S be any point in the plane of ABC, and H, K, L, as before, the mid points of BO, CA, AB. Let CZ || to SL and BY || to SK meet AB in Z, CA in Y, and each other in 0. Let AO meet BC in X. As before, let U, V, W be the mid points of OA, OB, OC. Obviously OVHW, OWKU, OULV are parallelograms, and SK || to OB II to UL"\ QT TTlr . „ SL |! to OC || to UK! •"• SLUK » .•. SHWK and similarly SHVL are parallelograms. Thus SHE=KW = U0, i.e. the lines through A, B, C parallel to SH, SK, SL are concurrent. Downloaded from https://www.cambridge.org/core. 01 Oct 2021 at 06:15:37, subject to the Cambridge Core terms of use . 132 We have also SO, UH, VK, LW bisecting one another in the same point M. Now UVWMO and ABCSO are homothetic, the ratio of their linear dimensions being 1 : 2. Hence if any figure be described about ABC, a homothetic figure of half of its linear dimensions can be described about UVW. If the circumscribing figure be a central curve of any sort, with S as centre, then the homothetic curve round UVW will have M as centre, and will therefore pass through HKL. If the central curve be a conic, so that supplementary chords are parallel to conjugate diameters and conversely, then XYZ will also lie on the conic UVWHKL. For SH and BC being conjugate directions as to the conic through ABC, they are so as to the conic UVWHKL. Thus HX, UX, drawn through the extremities of a diameter HU, are parallel to a pair of conjugate diameters, and therefore X lies on the conic. Similarly Y and Z. Every line from O to a point of the smaller conic meets the larger at double the distance. We note the special cases of the points X, Y, Z, H, K, L. Thus the properties of the nine-points conic investigated by Dr Pinkerton in his paper in Vol. XXIV. have been arrived at by Wilkinson's method, with the additional aid of the elements of the theory of homothesis, and the proposition about supplementary chords and conjugate diameters of a conic. S3. The whole of the preceding theory can be generalized further by first generalizing the fundamental propositions (i)...(viii) on which it is based. We define certain phrases. Given a fixed line a, if H is the harmonic conjugate as to BC of its intersection with a, we shall say that H "bisects BC as to a." If AB and CD are concurrent in a and also AC and BD, we shall say AB = CD and AC = BD. Hence the symbol = refers only to segments of lines which are concurrent in a. When a is the line at infinity, this meaning reduces to that previously assigned to the symbol =, and the fundamental Downloaded from https://www.cambridge.org/core. 01 Oct 2021 at 06:15:37, subject to the Cambridge Core terms of use . 133 property. "If AB = CD and CD = EF, then AB = EF," required to make the preceding proofs apply, is obviously a special case of Desargues' Theorem applied to the perspective triangles ACE, BDF. In place of the word parallel we shall in the generalized theory substitute concurrent in a. For our fundamental propositions (i)-(viii) in the generalized theory we have: (i) and (ii) are immediate results of the extended definitions, (iii) If AB^CD, then AD and BC "bisect one another as to a." This is the harmonic property of the complete quadri- lateral, (iv) If AD and BC " bisect one another as to a," then AB = CDand AC = BD. (v) If D "bisects AB as to a" and E "bisects AC as to a," then BC = 2DE. (iv) and (v) are cases of the proposition that if two harmonic ranges have one pair of corresponding points coincident, the joins of the other corresponding pairs are concurrent. (vi) If D "bisects AB as to a" and DE and BC are concurrent in a, E being in AC, then E " bisects AC as to a." This is an application of the theorem that if one transversal of a pencil gives a harmonic range, so does any other, (vii) and (viii) do not occur in the generalized theory. The generalised theory begins with H, K, L " bisecting as to a " the sides BC, CA, AB, and BY, CZ concurrent in a with SK, SL respectively, S being any point. O is the intersection of BY, CZ. U, V, W " bisect" OA, OB, OC ''as to a." (See Fig. 2.) LU and BO are concurrent in a by (v). .•. LU and SK are concurrent in a. Similarly SL and UK „ = 0U. .•. AX and SH are concurrent in a. It follows that: If lines be drawn through A, B, C respectively, concurrent in a with SH, SK, SL respectively, where H, K, L " bisect " BC, CA, AB "as to a," these lines will be concurrent in a point 0. Downloaded from https://www.cambridge.org/core. 01 Oct 2021 at 06:15:37, subject to the Cambridge Core terms of use . 134 Downloaded from https://www.cambridge.org/core. 01 Oct 2021 at 06:15:37, subject to the Cambridge Core terms of use . 135 The eight points 0, S, U, V, W, H, K, L, are such that OS, HU, KV, LW all " bisect one another as to a" in the same point M.